(I am working out of Hatcher. This is theorem 2.13.)
I am brewing up some confusion about the long exact sequence of the homology groups of $A \subset X$ and $X / A$. (For (X,A) a good pair, which is to say, that A is a closed subspace of X contained in a open set V that deformation retracts to A.)
I understand how a short exact sequence of chain complexes produces (naturally) a long exact sequence in homology, and also how one can use that and excision to show that $\tilde H_n(X/A) \cong H_n(X/A,A/A) \cong H_n(X,A)$ ($\tilde{}$ denotes the reduced homology.)
What I don't understand is how the following long exact sequence is derived from these facts:
$\ldots \to \tilde H_n(A) \to \tilde H_n(X) \to \tilde H_n(X/A) \to \tilde H_{n-1}(A) \ldots \to \tilde H_0(X/A) \to 0$
Of course, I see how this result follows above the zeroth homology, just by substituting the previously described identity into the long exact sequence of the pair. At the zeroth homology I am confused. Is it somehow the case that $\tilde H_n(X/A) \cong \tilde H_n(X,A)$ - that would fix this, but it seems likely to be false... (since reducing homology removes one copy of $Z$ from the zeroth homology).
Can someone help me?